For practical engineering applications, the use of proper amount of significant figures when quoting numerical values is not as rigorous as it would be in the pure sciences; however, it is still very important that we understand and represent the accuracy and precision of our values in engineering. One of the worst sins in engineering is pretending that you know more than you do. An important aspect of practical engineering is dealing with uncertainty. To say that the moment resistance of a beam is $123.4324\mathrm{\,kNm}$ (seven significant figures), is very different than saying that it is $120\mathrm{\,kNm}$ (two significant figures).The first number suggests that we know the resistance of the beam to a high level of accuracy, which is extremely unrealistic.

Of course, as you will know from a general knowledge of significant figures, if I multiply two numbers, and one has 3 significant figures, and the other has five significant figures, then the resulting number can only be known to three significant figures (the lessor of five and three). So, even if I know one of my numbers to five significant figures (for example, the length in mm) it will usually be multiplied by a number that has much less precision (for example the elastic modulus or material strength). For most materials that we use, we only really know the elastic modulus or material strength to at most two significant figures. On the loading side, even if we have a very good idea of the mass of a certain object, we only know the acceleration due to gravity $g$ to at most three significant figures ($9.80\mathrm{\,m/s^2}$ or $9.81\mathrm{\,m/s^2}$). Since all loads due to gravity are based on $F=mg$, there is no point using any more than three significant figures for those loads.

## Slide-Rule Accuracy

To simplify significant figures for a civil engineering context, engineers will often use a method called *slide-rule accuracy*. Instead of keeping rigorous track of our significant figures for each calculation, this method acknowledges that we know most numbers to at most three to four significant figures. The method is based on the accuracy of slide-rules, which are analog calculators widely used by engineers before the advent of the modern electronic calculator. The slide rule basically consisted of different rulers which are joined together and can slide relative to each other. Calculations are performed by lining up tick marks on adjacent rulers. Since these slide rules were read manually (by the human eye) and you can only fit so many tick marks on a $15$ to $30$ centimetre ruler, they typically had an accuracy of between three to four significant figures. For numbers starting with a one, there may be four significant figures (e.g. $1.234$), but for numbers starting with something else, there may only be three significant figures (e.g. $0.234$).

Therefore, for slide-rule accuracy, you keep four significant figures if the number starts with a $1$ and three significant figures otherwise. See Table 1.1 for some examples of slide rule accuracy.

Raw Value | Slide Rule Accuracy |
---|---|

$4.5896$ | $4.59$ |

$45\,896$ | $45\,900$ |

$45\,896\,543$ | $45\,900\,000$ |

$45\,896\,543$ | $4.59\times 10^7$ |

$0.000\,458\,96$ | $0.000\,459$ |

$0.000\,458\,96$ | $4.59 \times 10^{-4}$ |

## Engineering Notation

In addition to using slide rule accuracy, engineers also find it convenient to use *engineering notation* when writing large or small numbers. In structural analysis, we often deal with numbers that are on the order of $10^6$ or $10^9$. In scientific notation, you would show these types of numbers as $x \times 10^n$ where $x$ is a number that only has one whole digit before the decimal such as $1.345\times 10^5$ or $5.2 \times 10^{-2}$ (but not $234.2 \times 10^{4}$ or $0.23 \times 10^{9}$). In engineering, this type of notation may be awkward because it makes it difficult to add up numbers quickly. For example, it is not very straight forward to calculate $1.34 \times 10^{4}$ plus $3.6 \times 10^{3}$? You would probably need to use scientific notation on your calculator to solve it confidently. In structural analysis, this is a common problem and introduces an extra opportunity for mistakes in calculations.

To simplify these kinds of problems, in engineering we often group large numbers such that their exponent is a multiple of three (i.e. $10^3$, $10^6$, $10^9$, $10^{-3}$, etc...). This engineering notation is convenient because our units also tend to be grouped in the same kinds of multiples of $10^3$ such as N, kN, MN (newtons, kilonewtons, meganewtons), or Pa, kPa, MPa (pascals, kilopascals, megapascals). This type of notation reduces the chances for calculation error. The previous example would change to $13.4 \times 10^{3}$ plus $3.6 \times 10^{3}$. This is a subtle but important difference. To calculate the answer for this problem, we can easily just add $13.4$ and $3.6$ and then tack the exponent onto the end (which can be done in your head for this problem). This also simplifies unit conversions. Some examples of engineering notation versus scientific notation are shown in Table 1.2.

Scientific Notation | Engineering Notation |
---|---|

$4.59 \times 10^{3}$ | $4.59 \times 10^{3}$ |

$4.59 \times 10^{4}$ | $45.9 \times 10^{3}$ |

$4.59 \times 10^{11}$ | $459 \times 10^{9}$ |

$4.59 \times 10^{-5}$ | $45.9 \times 10^{-6}$ |